The binomial distribution that we discussed previously assumes replacement. That is, the
probabilities p and 1-p don’t change as “successes” or “failures” are witnessed.
That is, if we see a success, we theoretically put that success back into the population
so that our likelihood of successes doesn’t change, we “replace” that observation
back into the population. We assume that p and 1-p are probabilities that represent averages
probabilities found from a large number of observations that are sufficiently independent
of each other.
With the next distribution that we’re going to discuss, the hypergeometric distribution,
we address situations where the independence assumption isn’t met. That is, if we view
a success in an observation, then we have one fewer success that we can view in the
remaining sample. The hypergeometric distribution is good for describing a finite, and probably
small, population, where we know that have the population falls into one category (call
them “successes”) and the other half falls into the complement of that category (call
them “failures”). As such, the hypergeometric distribution is often said to assume no replacement.
Say that we’re assembling a committee of size five selected at random from three chemists
and six physicists. Other than sounding like a good time, what else do we know about this
committee? We know that it is being drawn from a finite group, and we know that the
group is made up of two categories: chemists and physicists. In the binomial problem, we
knew that some percentage generally fell into one category, therefore the complement fell
into the other category. But here, we know exactly how many are in each category in this
finite population. We also know that the members of the committee are not independent. For
example, if we add one chemist to the committee, we have one fewer chemist remaining to be
chosen, therefore the likelihood of choosing another chemist changes.
Let’s generally define the random variable Y as the number of successes in a number of
trials n. Capital N is the population size – we’re picking a number little n out
of this population. We know that the population is made up of two categories: r of the N are
successes, therefore the remaining N-r are failures. Little y is the number of successes
that we found in our group little n, therefore the remaining observations in our group are
n minus y failures. The probability that random variable Y takes on the value little y is
described with a classical probability: the number of ways we can witness little y successes,
and therefore, n-y failures, in the numerator, and the total number of ways we can choose
the group of n regardless of the numbers of successes and failures in the denominator.
We use combinations to find this probability. r choose y times (N minus r) choose (n minus
y) in the numerator. This gives us the number of ways to get a group of n with y successes.
Then N choose n, the number of ways to pick a group of n out of N, in the denominator.
Using the language of events, another way to think about this probability is with an
intersection statement in the numerator.
Back to our committee of five selected at random from three chemists and six physicists.
We define our random variable X as the number of chemists on the committee of size five.
What is the probability that we select a committee that has 2 chemists? We have to choose 2 chemists
out of 3 possible chemists. Then naturally we have to have 3 more members on the committee,
and they must be physicists, so we have to choose 3 physicists from the 6 physicists.
In the denominator, we count the ways we can select a committee of 5 across all 3+6 = 9
choices. This probability is 0.4762. Note that the possible realizations of random variable
X includes only 0, 1, 2, and 3. And we can use the hypergeometric distribution function
to calculation the probability of each of those realizations. We see from the distribution
that the most likely realization is X = 2: we are most likely to see a committee of size
five made up of 2 chemists and 3 physicists. Recall again the properties of probability:
each realization has a probability between 0 and 1, inclusive of 0 and 1, and the sum
of probabilities of all the realizations must be 1 (or maybe a little off due to rounding
errors).
The expected value and variance are a little difficult to derive, so we’ll just show
the punchline. The expected value is n times r divided by N. In our committee example,
the expected value of random variable X would be 5 times 3 divided by 9, or 1.67 chemists:
on average, we’d expect to see 1.67 chemists on the committee. That is, if we simulated
this process, choosing a committee of size 5 from 3 chemists and 6 physicists over and
over a bunch of times, we’d see 1.67 chemists on average.
The variance is much less intuitive. The variance of the number of chemists on the committee
would be 0.56 chemists squared, note, that’s different from square chemists. And the standard
deviation would be 0.75 chemists.
To recap, like the binomial distribution, we’re interested in counting the number
of observations that fall into one of two categories. Unlike the binomial distribution,
we assume no replacement occurs: we have a finite population, we know the exact makeup
of that population with respect to the two categories, and we calculate the probability
of drawing a particular number of one category in a group of size n from the population.